Question

A yo-yo (as shown) falls under gravity. Assume that it falls straight down, unwinding as it goes. Find the Lagrange equation of motion. Hints: The kinetic energy is the sum of the translational energy $\frac{\mathbf{1}}{\mathbf{2}}\mathbf{m}{\dot{\mathbf{z}}}^{\mathbf{2}}$and the rotational energy $\frac{\mathbf{1}}{\mathbf{2}}\mathbf{I}{\dot{\mathbf{\theta }}}^{\mathbf{2}}$ where $\mathbf{I}$ is the moment of inertia. What is the relation between $\dot{\mathbf{z}}$ and $\dot{\mathbf{\theta }}$ ?Assume the yo-yo is a solid cylinder with inner radius and outer radius $\mathbf{b}$.

Short answer

The relation between $\dot{\mathrm{z}}$and $\dot{\mathrm{\theta }}$ is $\left(\mathrm{m}+{\mathrm{la}}^{-2}\right)\ddot{\mathrm{z}}+\mathrm{mg}=0$.

Step-by-step solution

Given Information.

The given values are the kinetic energy is the sum of the translational energy $\frac{1}{2}\mathrm{m}{\dot{\mathrm{z}}}^2$and the rotational energy $\frac{1}{2}\mathrm{I}{\dot{\mathrm{\theta }}}^2$.

Step 2: Meaning of the Lagrange equations.

The Lagrange equations are used to construct the equations of motion of a solid mechanics issue in matrix form, including damping.

Find the relation between z˙ and θ˙ .

Since they are not independent. Assuming that at$\mathrm{z}=0$

the coordinate$\mathrm{\theta }=0$as well,

$$\begin{gathered} \mathrm{z}=-\mathrm{a\theta } \\ \dot{\mathrm{\theta }}=\frac{-\dot{\mathrm{z}}}{\mathrm{a}} \end{gathered}$$

rotational energies, therefore:

The potential energy has the following form:

$\mathrm{V}=\mathrm{mgz}$

And therefore the Lagrangian is:

Observe the Euler equation for$\mathrm{z}$degree of freedom. The Euler equations reads:

$\frac{\mathrm{d}}{\mathrm{dt}}\frac{\partial \mathrm{L}}{\partial \mathrm{z}}-\frac{\partial \mathrm{L}}{\partial \mathrm{z}}=0$

Calculate the required derivatives.

Use all of the equations above,

$\left(\mathrm{m}+{\mathrm{la}}^{-2}\right)\ddot{\mathrm{z}}+\mathrm{mg}=0$

For a cylinder with inner radius $\mathrm{a}$and outer radius $\mathrm{b}$the moment of inertia is:

$\mathrm{I}=\frac{1}{2}\mathrm{m}\left({\mathrm{a}}^2+{\mathrm{b}}^2\right)$

This can be inserted into the Euler equation but is not necessary.

Therefore, the relation between $\dot{\mathrm{z}}$and $\dot{\mathrm{\theta }}$ is$\left(\mathrm{m}+{\mathrm{la}}^{-2}\right)\ddot{\mathrm{z}}+\mathrm{mg}=0$.